Problem: A square is inscribed in the ellipse
\[\frac{x^2}{3} + \frac{y^2}{6} = 1,\]so that its sides are parallel to the coordinate axes.  Find the area of the square.

[asy]
unitsize(1 cm);

draw(xscale(sqrt(3))*yscale(sqrt(6))*Circle((0,0),1));
draw((sqrt(2),sqrt(2))--(-sqrt(2),sqrt(2))--(-sqrt(2),-sqrt(2))--(sqrt(2),-sqrt(2))--cycle);
draw((-2,0)--(2,0));
draw((0,-3)--(0,3));
[/asy]
Solution: By symmetry, the vertices of the square are $(\pm t, \pm t)$ for some positive real number $t.$  Then
\[\frac{t^2}{3} + \frac{t^2}{6} = 1.\]Solving, we find $t^2 = 2.$  Then $t = \sqrt{2}.$

The side length of the square is then $2t = 2 \sqrt{2},$ so its area is $(2 \sqrt{2})^2 = \boxed{8}.$